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Edexcel GCSE Computer Science (1CP2) Exam Questions

Build a practice paper from 604 exam-style questions, each with its mark scheme and example answers. Print it, or answer it on paper and then open the mark schemes. Or browse the questions by topic.

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4 questions · 5 marks

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1Multiple choice1 markFetch Decode Execute Cycle Swap Remove

Identify the CPU component that decodes instructions during the fetch-decode-execute cycle.

[1 mark]
  • AArithmetic logic unit
  • BControl unit
  • CMain memory
  • DRegister
Mark scheme
  • B Control unit (1)
  • The only correct answer is B. The correct answer is control unit.
  • A is not correct because the arithmetic logic unit performs arithmetic and logical operations, not decoding
  • C is not correct because main memory stores instructions and data, not decoding
  • D is not correct because a register holds data or addresses temporarily, not decoding
Example answers and how they were marked
Full marks1/1

B Control unit

Why: Correct option: the control unit decodes instructions (1).

No marks0/1

D Register

Why: Near-miss: a register holds data or addresses temporarily, it does not decode instructions (0).

No marks0/1

A Arithmetic logic unit

Why: Misconception: the arithmetic logic unit performs arithmetic and logical operations; it does not decode (0).

2Multiple choice1 markCharacter Sets Swap Remove

Characters are encoded using 7-bit ASCII.

Identify the number of different characters that can be represented using 7-bit ASCII.

[1 mark]
  • A64
  • B127
  • C128
  • D256
Mark scheme
  • C 128 (1)
  • The only correct answer is C. The correct answer is 128.
  • A is not correct because 6 bits give only 64 different combinations
  • B is not correct because 127 is the largest 7-bit value, not the number of characters that 7 bits can represent
  • D is not correct because 256 characters need 8 bits, as used by 8-bit extended ASCII
Example answers and how they were marked
Full marks1/1

C 128

Why: Selects the only correct option (1): 7 bits give 2 to the power of 7, which is 128 different characters.

No marks0/1

D 256

Why: Near miss scoring 0: 256 is the number of characters that 8 bits give in extended ASCII, not 7-bit ASCII.

No marks0/1

B 127

Why: Misconception scoring 0, the off-by-one error the examiners' report describes: 127 is the largest 7-bit value, but 7 bits give 128 different characters.

3Short recall2 marksStored Program Computers Swap Remove

The central processing unit (CPU) contains an arithmetic logic unit (ALU).

State two types of operation carried out by the ALU.

[2 marks]
Mark scheme

Award one mark for any of the following up to a maximum of two marks:

  • An arithmetic operation, e.g. addition/subtraction/multiplication/division (1)
  • A logical operation, e.g. comparing two values (is A > B?) / using AND, OR, NOT (1)

Do not accept decoding instructions, fetching instructions or controlling the timing of the CPU, as these are carried out by the control unit.

Example answers and how they were marked
Full marks2/2

It carries out arithmetic operations such as addition and subtraction, and logical operations such as comparing two values to see if one is greater than the other.

Why: Arithmetic operation (1) and logical operation (1).

Some marks1/2

It adds numbers together and works out the result.

Why: Arithmetic operation, addition (1). No logical operation.

No marks0/2

It decodes the instructions that have been fetched from main memory.

Why: Decoding and fetching are the control unit's role; the mark scheme's do-not-accept line covers this.

4Short recall1 markUnits & Conversion Swap Remove

A file has a size of 2 kibibytes.

State the number of bytes in the file.

[1 mark]
Mark scheme

Award one mark for:

  • 2048 (1)
  • 2 × 1024 (1)
  • 2^11 (1)

Allow 2048 bytes.

Do not accept 2000, which is the number of bytes in two decimal kilobytes.

Do not accept 1024, which is the number of bytes in one kibibyte.

Do not accept 16, which is the number of bits in two bytes.

Example answers and how they were marked
Full marks1/1

2048

Why: 2048 is 2 × 1024, the number of bytes in 2 kibibytes (1).

No marks0/1

2000

Why: 2000 uses the decimal kilobyte, not the binary kibibyte; the mark scheme does not accept it.

No marks0/1

16

Why: 16 is the number of bits in two bytes, confusing bits with bytes; the mark scheme does not accept it.

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